Answer:
Explanation:
The surface area of a star estimated by the energy emitted per sq meter yields the overall luminosity, which can be represented mathematically as:
[tex]L= 4 \pi R^2 \sigma T^4 --- (1)[/tex]
where;
L ∝ R²T⁴
and;
R = radius of the sphere
σ = Stefans constant
T = temperature
Also; The following showcase the relationship between flux density as well as illuminated surface area as:
[tex]F = \dfrac{L}{A}[/tex]
where
A = 4πd² and L ∝ R²T⁴
[tex]F = \dfrac{R^2T^4}{4 \pi d^2} \\ \\ F \alpha \dfrac{R^2T^4}{ d^2} --- (2)[/tex]
Given that:
distance d₁ = 20 pc
Then, using equation (2)
[tex]F_1 \ \alpha \ \dfrac{R^2_1T^4_1}{ d^2_1}[/tex]
However, we are also being told that there is a temp. drop by a factor of 3;
So, the final temp. [tex]T_2 = \dfrac{T_1}{3}[/tex]; and the final radius is [tex]R_2 = 100R_1[/tex] since there is increment by 100 folds.
Now;
[tex]F_2 \ \alpha \ \dfrac{R^2_2T^4_2}{ d^2_2}[/tex]
SInce;
[tex]F_1 = F_2[/tex]
It implies that:
[tex]\dfrac{R^2_1T^4_1}{ d^2_1 } = \dfrac{R^2_2T^4_2}{ d^2_2} \\ \\ d_2 = \sqrt{\dfrac{R_2^2T_2^4}{R_1^2T_1^4}}(d_1)[/tex]
Replacing all our values, we have:
[tex]d_2 = \sqrt{\dfrac{(100R_1)^2 \times (\dfrac{T_1}{3})^4}{R_1^2T_1^4}}(20 ) \\ \\ d_2 = \sqrt{\dfrac{(100)^2 }{3^4}}(20 ) \\ \\ d_2 = \sqrt{\dfrac{(100)^2 }{3^4}}(20 ) \\ \\ d_2 =222 \ pc[/tex]
